# Coefficent of Friction Question

1. Oct 15, 2007

### UFGators35

1. The problem statement, all variables and given/known data

A 1200 kg car traveling at 13 m/s is able to skid to a stop over a distance of 30 meters. What must be the coefficient of friction between the road and the tires for this to be possible?

m= 1200kg
Vo= 13 m/s
V= 0 m/s
x= 30 m

2. Relevant equations

I think this is kinetic friction so:

fk=µkFn

3. The attempt at a solution

I know I want to solve for µk but we haven't been over this in class and I am not really sure how to solve for it. I also don't know how I arrive at fk and Fn from the info I was given.

2. Oct 15, 2007

### hage567

Do you know how to find the acceleration given the info in the question?

Do you know what the normal force is?

Do you know Newton's second law?

3. Oct 15, 2007

### UFGators35

The acceleration:
V^2x= Vo^2x + 2ax(x)
a= -Vo^2x / 2x
a= -169 / 60
a= -2.8166667 m/s^2

Normal Force:
ΣF=ma
ΣF= 1200 * -2.8166667
ΣF= -3380 N

4. Oct 15, 2007

### hage567

Your first part for the acceleration is right. But the normal force is not what you have in the second part. The normal force is the reaction force of the car on the road. It acts perpendicular to the surface of the road, and has a magnitude of Fn = mg. You need to put that into your equation for the frictional force that you have in your first post. Since you know m, g and have found m*a, you can solve for $$\mu_k$$.